Difference between revisions of "User:Nik"
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| + | Let | ||
| + | $$ \Phi \in C^\infty(\mathbb{R}^n) $$ | ||
| + | have the property that the system | ||
| + | $$p_i = \frac{\partial}{\partial q_i} \Phi (q, \widetilde{q}) $$ | ||
| + | has a unique smooth solution | ||
| + | $$ \widetilde{q} = \widetilde{q} (q,p) $$ | ||
| + | |||
| + | Define | ||
| + | $$ \widetilde{p_i}(q,p) = - \frac{\partial}{\partial \widetilde{q_i}} \Phi (q, \widetilde{q}) | _{\widetilde{q}= \widetilde{q}(q,p)} $$ | ||
| + | |||
| + | Let $$ \{\cdot,\cdot\} $$ be the Poisson bracket, such that | ||
| + | $$ \{f,g\} = \sum_{j=1}^n \frac{\partial f}{\partial q_j} \frac{\partial g}{\partial p_j} - \frac{\partial f}{\partial p_j} \frac{\partial g}{\partial q_j} $$ | ||
| + | |||
| + | Show that: $$\\$$ | ||
| + | i) | ||
| + | $$ \{\widetilde{q_i} (q,p), \widetilde{q_j} (q,p)\} = \{\widetilde{p_i} (q,p), \widetilde{p_j} (q,p)\} = 0 $$ | ||
| + | |||
| + | ii) | ||
| + | $$ \{\widetilde{q_i} (q,p), \widetilde{p_j} (q,p)\} = \delta_{ij} $$ | ||
| + | |||
| + | |||
=== Solution for i) === | === Solution for i) === | ||
| + | === Solution for ii) === | ||
Revision as of 09:19, 28 December 2014
Let $$ \Phi \in C^\infty(\mathbb{R}^n) $$ have the property that the system $$p_i = \frac{\partial}{\partial q_i} \Phi (q, \widetilde{q}) $$ has a unique smooth solution $$ \widetilde{q} = \widetilde{q} (q,p) $$
Define $$ \widetilde{p_i}(q,p) = - \frac{\partial}{\partial \widetilde{q_i}} \Phi (q, \widetilde{q}) | _{\widetilde{q}= \widetilde{q}(q,p)} $$
Let $$ \{\cdot,\cdot\} $$ be the Poisson bracket, such that $$ \{f,g\} = \sum_{j=1}^n \frac{\partial f}{\partial q_j} \frac{\partial g}{\partial p_j} - \frac{\partial f}{\partial p_j} \frac{\partial g}{\partial q_j} $$
Show that: $$\\$$ i) $$ \{\widetilde{q_i} (q,p), \widetilde{q_j} (q,p)\} = \{\widetilde{p_i} (q,p), \widetilde{p_j} (q,p)\} = 0 $$
ii) $$ \{\widetilde{q_i} (q,p), \widetilde{p_j} (q,p)\} = \delta_{ij} $$
